Question

Determine the constants \(b\) and \(c\) such that the line tangent to \(f(x)=x^{2}+b x+c\) at \(x=1\) is \(y=4 x+2\).

Short answer

Question: Determine the constants b and c such that the tangent to the quadratic function f(x) = x^2 + bx + c at x = 1 is y = 4x + 2. Answer: The constants are b = 2 and c = 3, so the function is f(x) = x^2 + 2x + 3.

Step-by-step solution

Find the derivative of the given function

To find the slope of the tangent line to the function \(f(x)=x^2+bx+c\), we first need to find its derivative, \(f'(x)\). Using the power rule, we have: \(f'(x) = 2x + b\) This expression represents the slope of the tangent line to the curve at any point \(x\).

Determine the slope of the tangent at \(x = 1\)

To find the slope of the tangent line at the point \(x = 1\), we need to replace \(x\) with 1 in the derivative that we found in the previous step: \(f'(1) = 2(1) + b = 2 + b\)

Determine the slope of the given tangent line

We are given that the tangent line is \(y = 4x + 2\). The slope of this line is 4, as it is the coefficient of the \(x\) term.

Match the slopes of the tangent and use it to find the value of \(b\)

Since the slopes of the tangent line and the derivative at the point of tangency should be equal, we have: \(2 + b = 4\) Solving for \(b\), we get \(b = 2\).

Find the value of \(c\)

Thanks to the point-slope form, we know that the coordinates of the point of tangent must satisfy both the function \(f(x)\) and the tangent line. Thus: \(f(1) = (1)^2 + 2(1) + c\) and \(y = 4(1) + 2\) Comparing the two expressions, we get: \(1 + 2 + c = 4 + 2\) Solving for \(c\), we find that \(c = 3\).

Write down the final function

Now we have determined both constants \(b\) and \(c\). The function with the tangent \(y = 4x + 2\) at \(x = 1\) is: \(f(x) = x^2 + 2x + 3\)

Key concepts

Derivative

Understanding the concept of derivative is crucial in calculus because it helps us find the rate of change at any given point. In simpler terms, it tells us how a function is changing at any point on its curve. When you think of a car's speed, for instance, the speedometer shows your speed at that exact moment—this is like a derivative.
The derivative of a function at a specific point gives you the slope of the tangent line at that point, indicating how steep the curve is.

• The derivative of a function is often denoted by symbols such as \(f'(x)\), \(dy/dx\), or \(D(f(x))\).
• Derivatives can be used to find slopes, rates of change, and can even help in optimizing problems in real-world scenarios.

To find a derivative, mathematicians use various rules including the chain rule, product rule, quotient rule, and the power rule which we'll discuss next.

Power Rule

The power rule is one of the simplest and most commonly used rules in differentiation, thanks to its straightforward application. If you have a function that is a power of \(x\) like \(x^n\), the power rule states that its derivative will be \(nx^{n-1}\). This means you bring down the power as a coefficient and subtract one from the power.
The power rule works great with polynomials, where each term can be differentiated individually.

• For example, given \(f(x) = x^2\), its derivative using the power rule is \(f'(x) = 2x\).
• For a general polynomial like \(f(x) = ax^b\), the derivative is simply \(f'(x) = abx^{b-1}\).

In the original exercise, this rule was used to find the derivative of the function \(f(x) = x^2 + bx + c\). By applying the power rule, the derivative \(f'(x) = 2x + b\) was calculated.

Slope of the Tangent Line

The slope of a tangent line to a curve at a given point is vital in understanding how the curve behaves at that specific point. A tangent line is a straight line that "just touches" the curve at that point without crossing it.
The slope of the tangent line, numerically, is the same as the derivative of the function at that particular point.

• This slope is positive if the curve is increasing and negative if it is decreasing.
• At a point where the curve flattens out, such as a peak or valley, the slope (and thus the derivative) is zero.

From the initial problem, the slope of the tangent line was given as 4, hence by equalizing \(f'(1) = 2 + b\) with 4, we found that \(b = 2\). This process shows the direct use of derivatives in determining slopes.

Point-Slope Form

The point-slope form is a practical way to write the equation of a line when you know a point on the line and its slope. The general formula for point-slope form is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a specific point on the line.
This is particularly useful in problems involving tangents to curves.

• In our problem, knowing the tangent line as \(y = 4x + 2\) gives us both the slope \(m = 4\) and a point \((1, 4(1)+2)\).
• With this information, you can determine constants or verify the structure of the original function, as shown in the step-by-step solution.

By placing the calculated tangent point into the function \(f(x) = x^2 + bx + c\) and knowing that point is also on the line \(y = 4x + 2\), we validated and found our unknowns \(b\) and \(c\), ensuring the curve and line intersect correctly.